Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 78

Determine \(u_{\mathrm{m}}, \bar{u},\) and \(u_{\mathrm{rms}}\) for a group of ten automobiles clocked by radar at speeds of 38,44,45,48,50 \(55,55,57,58,\) and \(60 \mathrm{mi} / \mathrm{h},\) respectively.

Short Answer

Expert verified
The mean speed of the vehicles \(\bar{u}\) is 51.0 mi/h, the mode of the speeds \(u_{\mathrm{m}}\) is 55 mi/h, and the root mean square speed \(u_{\mathrm{rms}}\) is 52.39 mi/h.

Step by step solution

01

Calculation of Mean \(\bar{u}\)

The mean \(\bar{u}\) for the speed of cars is defined as the sum of all speed values divided by the number of values. That is: \(\bar{u} = \frac{1}{n} \sum_{i=1}^{n} u_i\). So, add up the speeds: 38 + 44 + 45 + 48 + 50 + 55 + 55 + 57 + 58 + 60 and then divide by 10.
02

Calculation of Mode \(u_{\mathrm{m}}\)

The mode \(u_{\mathrm{m}}\) is the speed value that appears most frequently in the dataset. If you look at the dataset, it shows that the speed value 55 appears twice. Therefore the mode of this set is 55.
03

Calculation of Root Mean Square \(u_{\mathrm{rms}}\)

The root mean square \(u_{\mathrm{rms}}\) for the speed of cars is defined as the square root of the mean of the squares of all the values. That is: \(u_{\mathrm{rms}} = \sqrt{\frac{1}{n} \sum_{i=1}^{n} u_i^2}\). So square each speed: \(38^2, 44^2, 45^2, 48^2, 50^2, 55^2, 55^2, 57^2, 58^2, \) and \(60^2\), then add these results together, divide by 10 and then take the square root of the result.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

According to the CRC Handbook of Chemistry and Physics (83rd ed.), the molar volume of \(\mathrm{O}_{2}(\mathrm{g})\) is \(0.2168 \mathrm{Lmol}^{-1}\) at \(280 \mathrm{K}\) and \(10 \mathrm{MPa}\). (Note: \(1 \mathrm{MPa}=\) \(\left.1 \times 10^{6} \mathrm{Pa} .\right)\)(a) Use the van der Waals equation to calculate the pressure of one mole of \(\mathrm{O}_{2}(\mathrm{g})\) at \(280 \mathrm{K}\) if the volume is 0.2168 L. What is the \% error in the calculated pressure? The van der Waals constants are \(a=1.382 \mathrm{L}^{2}\) bar \(\mathrm{mol}^{-2}\) and \(b=0.0319 \mathrm{L} \mathrm{mol}^{-1}\) (b) Use the ideal gas equation to calculate the volume of one mole of \(\mathrm{O}_{2}(\mathrm{g})\) at \(280 \mathrm{K}\) and \(10 \mathrm{MPa}\). What is the \% error in the calculated volume?

A 0.7178 g sample of a hydrocarbon occupies a volume of \(390.7 \mathrm{mL}\) at \(65.0^{\circ} \mathrm{C}\) and \(99.2 \mathrm{kPa}\). When the sample is burned in excess oxygen, \(2.4267 \mathrm{g} \mathrm{CO}_{2}\) and \(0.4967 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) are obtained. What is the molecular formula of the hydrocarbon? Write a plausible structural formula for the molecule.

A 2.00 L container is filled with \(\operatorname{Ar}(g)\) at 752 mm Hg and \(35^{\circ} \mathrm{C} .\) A \(0.728 \mathrm{g}\) sample of \(\mathrm{C}_{6} \mathrm{H}_{6}\) vapor is then added. (a) What is the total pressure in the container? (b) What is the partial pressure of \(\mathrm{Ar}\) and of \(\mathrm{C}_{6} \mathrm{H}_{6} ?\)

A particular gaseous hydrocarbon that is \(82.7 \%\) C and \(17.3 \%\) H by mass has a density of \(2.33 \mathrm{g} / \mathrm{L}\) at \(23^{\circ} \mathrm{C}\) and \(746 \mathrm{mm} \mathrm{Hg} .\) What is the molecular formula of this hydrocarbon?

The heat required to sustain animals while they hibernate comes from the biochemical combustion of fatty acids, such as arachidonic acid, \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\) What volume of air, measured at \(298 \mathrm{K}\) and \(1.00 \mathrm{atm}\) is required to burn \(2.00 \mathrm{kg} \mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2} ?\) Air is approximately \(78.1 \% \mathrm{N}_{2}\) and \(20.9 \% \mathrm{O}_{2},\) by volume. Other gases make up the remaining \(1.0 \%\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free